Symbolic Integration

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Symbolic Integration

• Lecture 14

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The problem of integration in finite terms

• The Freshman Calculus Problem
• The Artificial Intelligence Problem
• The Algebraic (Rational Function) Problem (Tobey,
  Horowitz, Rothstein, Trager)
• The Decision Problem (Liouville, Ritt,
  Rosenlicht, Risch, Cherry, Trager, Davenport,
  Bronstein, others)
• The Computer Algebra System problem.
• Extensions to definite integration, numerical
  integration, approximation by Taylor series,
  Laurent series, asymptotic series.

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The Freshman Calculus Problem. Find the flaws.

• Humans are intelligent
• MIT students are especially intelligent humans.
• Freshmen study integral calculus (this is in
  1960s before Advanced Placement).
• Therefore solving calculus problems requires
  intelligence.
• If a computer can solve calculus problems, then
  it is intelligent (An example of Artificial
  Intelligence).
• To solve these problems, imitate a freshman
  student.
• Future extensions do the rest of math, and the
  rest of AI.

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Freshman Calc ? AI James Slagles approach

• AND-OR trees for evaluation of difficulty of a
  path (do we abandon this route or plow on?)
• Game playing could be any complex task.

or
or
and
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Slagles program SAINT Symbolic Automatic
INTegrator

• ELINST elementary instance expression pattern
  matching (in assembler).
• Simplification program (in assembler).
• Used lisp prefix expressions ( x (sin x))
• First major lisp program (in Lisp 1.5).
• Took about a minute per problem (about the same
  time as a student!) running uncompiled on an IBM
  7090 (32k word x 36 bits/word).
• Could not do rational function integration (out
  of space.)

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The Algebraic (Rational Function) Problem

• Many attempts to do this right, e.g. Theses of
  R.Tobey, E. Horowitz, and with extensions, M.
  Rothstein, B. Trager.
• Main idea is to take a rational expression
• q(x)/r(x), express it as a polynomial proper
  fraction PQ/R with deg(Q)lt deg(R).
• Integrate P, a polynomial, trivially.
• Expand Q/R in partial fractions.

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Partial fractions trivialized

• From Q/R Find all the complex roots of R,
• a0, ...,an.
• Express Q/R as å ci/(x-ai) which integrates to
  å ci log(x-ai).
• multiple roots are simple c(x-a)-n ?
  c/(1-n)(x-a)1-n
• Can we always find ai? Complex roots can be
  found numerically. Can we find ci?
• It turns out that we can expand the summation and
  solve everywhere... with a big problem..

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an example of the general cubic...
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Whats the problem?

• If the denominator R s3x3s2x2s1xs0, then each
  of the roots ai look like this

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The answer is unmanageable even if we can compute
it.

• For the cubic, each ci looks something like this
• and so it is ugly. Often enough these
  expressions can be simplified, and so the task is
  to find a minimal algebraic extension in which
  the ci can be expressed. We DONT need to see the
  ai, but can use facts like a1a2a3 is the constant
  term of polynomial R we have neat forms for all
  symmetric functions of roots.

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For a general quartic, it gets worse.

• For order 5 or more there is in general no
  formula in radicals for the roots. That is, we
  cant expect to represent the ai for symbolic
  polynomials without extra notation (Elliptic Fns
  (not radicals) will do).
• We can find approximate numerical roots in C if
  the problem has a numeric denominator solely in
  Qx, but all bets are off symbolically.
• The solution is RootSum(f(x),poly,x)
  expressions where we have å f(ai) where ai are
  the roots of the polynomial poly.

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This trick works for numeric denominators too
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Rothstein/Trager/Rioboo/Lazard (et al), find the
ci easier

• Let B(x) be square free and of degree gt A(x),
• then s (A/B)dx ci log(vi)
• where ci, are the distinct roots of the
  polynomial R(c)Resultant(x,A(x)-cB(x),B(x)) and
  vigcd(A(x)-ciB(x),B(x)) for i 1,...n.
• We need Square-free computation and Resultant wrt
  x
• (Good reference Geddes/Czapor/Labahn)

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Squarefree isnt a problem

• Let the problem be sA(x)/B(x)n with
  deg(A)ltdeg(B).
• Since B itself is square free, its GCD with
  BdB/dx is 1. Thus we can compute c,e such that
  cBeB1.
• Now multiply through by A to get cABeBAA, and
  substitute in the original sA(x)/B(x)n to get
  s(cABeBA)/B(x)n . Now dividing through we get
• s cA/Bn-1 s eA (B/Bn) where the second term
  can be integrated by parts to lower the power of
  Bn in the denominator.
• (reminder s u dv uv-s v du. Let u eA,
  dvB/Bn,
• v1/((1-n)Bn-1), du(eA), so s v du comes out
  as ltsomethinggt/Bn-1 s ltsomethinggt/Bn-1 . Iterate
  until Bn B1 .)

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Resultant isnt a problem

• Well, actually it is something wed prefer not to
  do since it takes a while, but we have algorithms
  for it.
• Conclusion we can do rational function
  integration pretty well. But it took us into the
  mid-1980s to do it right.

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The decision problem

• Liouville 1809-1882 During the period
  (1833-1841) presented a theory of integration
  proved elliptic integrals cannot have elementary
  expressions.
• Various other writers advanced the subject in
  late 1800s
• J. Ritt (1948) Integration in Finite Terms
  Columbia Univ. Press
• M. Rosenlicht (AMM. 1972) Integration in Finite
  Terms
• R. Risch (1968,69) Solution to the problem of IFT
  unreadable
• B. Trager, J. Davenport, M.Bronstein
  implementation

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Rischs result
Theorem Let K be a differential field and f be
from K . Then an elementary extension of the
field K which has the same field of constants as
K and contains an element g such that g f
exists if and only if there exist constants
c1..cn from K and functions u0,..,un from K such
that f u0 åi1,...,nci(ui/ui) or gs fdx
u0 å i1..,n ci log(ui) note this allows
additional logs in the answer, but thats all.
The structure of the integral is specified. And
the notation g means the derivative of g.
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Whats a Differential Field?

• Equip a field with an additional operator D which
  satisfies identities parallel to the rules for
  differentiation of functions. These structures
  obviously include various fields of functions
  (e.g. the field of rational functions in one
  variable over the real field).
• The goal is to provide a setting to answer
  concretely such questions as, "Does this function
  have an elementary antiderivative?

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Formally, start with a field F and a derivation
map

• Derivation is a map of F into itself
• a ? a such that
• (ab) a b
• (ab)abab
• CONSEQUENTLY. (I.E. these are not new axioms)
• (a/b) (ab-ab)/b2 if a,b, 2 F and b¹ 0
• (an) nan-1a for all integers n
• 1 0 the constants F are a subfield of F

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Also

• If a, b are in a differential field F with a
  being nonzero, we call a an exponential of b or b
  a logarithm of a if b a/a.
• For algebraic extensions of F, there is a theorem
  that says that F is of characteristic zero and K
  is an algebraic extension field of F then the
  derivation on F can be extended uniquely to K.

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We must also deal with algebraic pieces

• Algebraic extensions z where p(z)0 in F
• Monomial extensions exp() and log(). Independent
  extensions are required e.g. log(x) and log(x2)
  dont cut it since (log(x2))-(2log(x)) is a
  constant (perhaps 0)

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examples

• sometimes we prefer arctan (if alt0 below),
  sometimes logs.
• Risch converts s xnsin x to exponentials
• and then cant do it..

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Extensions of the Risch algorithm

• Allow certain functions beyond log, exp such that
  the form of the integral still holds. For
  example, erf(x) (2/p)s0x exp(-t2)dt has the
  appropriate structure.

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The Risch Algorithm two steps back.

• There is a fallacy in claiming the Risch
  "algorithm" is an algorithm at all it depends,
  for solution of subproblems, on heuristics to
  tell if certain expressions are equivalent to
  zero.
• We know from Richardsons work that the
  zero-equivalence problem over a class of
  expressions much smaller than that of interest
  for integration is recursively unsolvable.
• But that is not the problem with most
  implementations of the Risch algorithm. They
  mostly have not been programmed completely,
  because, even assuming you can solve the
  zero-equivalence problem, the procedure is hard
  to program.

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The Risch Algorithm answers are not ncessarily
what you expect

• It is not nearly as useful as you might think,
  because it returns algebraic antiderivatives
  whose validity may be on a set of measure zero.
  Work by D. Jeffrey A. Rich (among others) on
  removing gratuitous discontinuities, is helpful.
• The Risch algorithm may also, in the vast
  majority of problems, simply say, after an
  impressive pause, nope. can't do it. There is no
  reasonable complexity analysis for the process,
  which is probably why certain authors ignore it.

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The Risch Algorithm answering wrong problem

• Also, most people are interested in definite, not
  indefinite integrals, at least once they've
  finished with Freshman calculus.
• And in cases where approximate solutions are
  easily obtained for the corresponding definite
  integral, the Risch algorithm may grind on for a
  long while and then say there is no closed form.
  Or if there is a closed form, if it is to be
  ultimately evaluated numerically, the closed form
  may be less useful than the quadrature formula!

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Nevertheless, the Risch Algorithm fascinates us

• The theory of algebraic integration, and its
  corresponding history is interesting.
• The solution to calculus problems can probably
  be presented this way.
• Its a good advertising slogan.
• The definite integration problem is also a
  classic in the applied mathematics literature
• It would seem that vast tables (20,000 entries)
  could be replaced with a computer program.
  (Actually Risch alg. is almost irrelevant here)

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It also eludes us.

• It is in some sense at the pinnacle of CAS
  algorithms it uses more mechanisms of a computer
  algebra system than nearly any other program.
• Rational function manipulation (GCD, partial
  fractions, factoring)
• Simplification in a differential field
  (algebraic, exponential/log, complex extensions)
• Solving certain ODEs
• J. Moses, J. Davenport, B. Trager, D. Lazard, R.
  Rioboo, A. Norman, G. Cherry, M. Bronstein,

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Go over examples of Risch in Moses paper

• Detailed understanding of Rischs original
  presentation is a challenge.

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How should one write an integration program?
(Moses)

• Quick solution to easy problems
• A collection of methods and transformations
• Radical methods
• (less known, table lookup)
• consider approximate or numerical approaches
• expand integrand in Taylor series, orthogonal
  polynomials, Fourier series, or asymptotic series
  or even approximate as rational functions.
• (for rational functions), approximate the roots
  of the denominator and do partial fraction
  expansion.
• do the whole task by quadrature (a well-studied
  area) treating the integrand as a black box
  capable only of returning a value at a point.
  difficulties with infinite range or nasty
  behaviors.

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Losing numerically

• Try integrating xsin(x) between 0 and 5000.
• Numerically, its tough unless you guess that it
  is periodic (certainly possible, but somewhat of
  a leap of faith if you just have a black box
  integrand).
• Symbolically it is 5000 cos (5000)sin(5000)
• Mathematica gives -774.31 (symbolic ? number) vs
  NIntegrate -- -210450)

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Harder Losing numerically, rapid oscillation..

• Or sin(1/x) between ? and 1.

integral is -CIx-1 x sinx-1 CI is cosine
integral ?x cos(t)/t dt, lesser known but still
tabulated function.
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Harder integrating through singularity
Examples take the difference or quotient of
functions whose singularities cancel, or which
have integrable singularities.
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Losing symbolically

• Sometimes, as mentioned earlier, the answer is
  possible but ugly. Consider this expression
  suggested by W. Kahan.. s1/(z641) dz integrated
  to
• Computer algebra systems get forms that are worse
  than this (try them!) and often contribute other
  monstrosities more easily integrated numerically